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Measuring Resistance

I have a basic series-parallel circuit, it is power by my DC power supply. From the Neg to R1, then to the parallel group of R2 & R3, then to the Pos. Question : Why is it when i measure the series resistance of R1 in the circuit with the power connected ( not turned on) I get a different reading than if I take it right out of the ciruit? Do I have to disconnect the power to read resistance because the power supply is actually in parallel to the series resistor of R1. that would mean the same is taking place with the parallel group but I haven't noticed since I was splitting that up anyhow to measure the resistance there. Is this something I should keep in mind for the future, disconnect the power line to the circuit before measuring resistance - even if only a series resistance.

Thank you in advance one and all... .. Bob

Reply to
Bob H.
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You can't measure resistance with the circuit live.

Graham

Reply to
Eeyore

And yes the power supply forms part of the cct when turned off.

Reply to
John G.

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I think your meter probably measures resistance by applying a regulated current through the resistor under test, and then taking a reading of the voltage across it. Or maybe it applies a voltage and measures current; either way, some power has to come from the meter and go through the resistor. If that's true, then having the circuit turned on and applying more voltage and current through the resistor would really screw up the reading. And you don't have to connect a battery to the circuit to mess up the reading. Consider connecting a resistor in place of the battery. You then have a different configuration altogether, with an added resistor in parallel with your network. So in general, you may not be able to trust measurements "in-circuit" of static values such as resistance. Dynamic measurements such as with an oscilloscope present the same concern; that's why o-scopes have such high input impedance.

Reply to
gearhead 

--
Because your power supply looks like a resistance when it\'s turned off.
(View in Courier)

          
        
   +--+-[R3[-+
   |+ |      | 
  [R] +-[R2]-+
   |-        |
   +----[R1]-+

 
JF
Reply to
John Fields

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